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Lesson Explainer: Solving Systems of Linear Equations Graphically Mathematics

In this explainer, we will learn how to solve a system of two linear equations by considering their graphs and identifying the point of intersection.

When we are asked to solve a system of equations, this means we are looking for a set of values for the variables that satisfy every equation. To see how we may solve a system of equations graphically, let’s consider the following example: 𝑦=π‘₯+1,𝑦=βˆ’2π‘₯+4.

We first notice that these equations are actually linear functions. Since there are two equations, we call this a system of two linear equations.

We want to find the π‘₯- and 𝑦-values that satisfy both equations. Let’s say that π‘₯=π‘Ž and 𝑦=𝑏 is a solution to both equations. We can then determine potential values for π‘Ž and 𝑏 graphically. We note that since 𝑏=π‘Ž+1, we know that the point (π‘Ž,𝑏) lies on the graph of the first equation. Similarly, since 𝑏=βˆ’2π‘Ž+4, the point (π‘Ž,𝑏) lies on the graph of the second equation. Therefore, (π‘Ž,𝑏) lies on both graphs, so it is a point of intersection between the graphs.

We can sketch the graphs of these equations by noticing that both equations represent straight lines in the coordinate plane. In fact, they are given in slope–intercept form. We sketch the first line with a 𝑦-intercept of 1 and slope 1 and the second line with a 𝑦-intercept of 4 and a slope of βˆ’2. This gives us the following.

We can read the coordinates of the point of intersection as (1,2). It is worth noting at this point we have reasoned that if (π‘Ž,𝑏) is a solution to the system of equations, then it is a point of intersection between the graphs. However, we also want to check that all points of intersection are solutions. We can do this by noting if (π‘Ž,𝑏) is a point of intersection, then (π‘Ž,𝑏) lies on the graph of each equation and hence must satisfy every equation; otherwise, it will not lie on its graph. Therefore, every point of intersection (if it exists) is a solution to the system of equations.

We can use this to conclude that since this is the only point of intersection, π‘₯=1, 𝑦=2 is the only solution to this system of equations.

We can also note that if there is no point of intersection between all of the graphs of the equations, then there are no values for the variables that satisfy all of the equations. This then gives us the following result.

Theorem: Solving a System of Equations Graphically

Every solution to a system of equations is a point of intersection of all of their graphs.

Every point of intersection on all of the graphs of the equations will be a solution to the system of equations.

If there are no points of intersection between all the graphs, then there are no solutions to the system of equations.

If we are working with a system of two linear equations, then we know that both equations can be graphed to give straight lines. We can then recall that there are three possibilities for the intersection of two straight lines in the plane.

  1. The lines intersect at exactly one point.
  2. The lines are distinct and parallel, so there are no points of intersection.
  3. The lines are coincident, so the lines intersect at every point on the lines.

Since the points of intersection are the solutions to the system of equations, these three possibilities tell us the number of solutions to these systems of equations. There are three possibilities for the number of solutions to a system of two linear equations; there are either 0,1, or an infinite number of solutions, depending on the lines. It is worth noting this also holds for more than two lines. We give each of these three scenarios a name.

Definition: Number of Solutions to a System of Linear Equations

If a system of linear equations has one solution, then there is a single point of intersection between the lines and we call this an independent system of linear equations.

If a system of linear equations has an infinite number of solutions, then all of the lines are coincident and we call this a dependent system of linear equations.

In both of the cases, we have at least one solution to the system, so we call these consistent systems of equations.

If a system of linear equations has no solutions, then there are no points of intersection between the lines and we call this an inconsistent system of equations.

Let’s use the graphical method to solve the following system of two linear equations: 𝑦=4π‘₯+3,2π‘₯+𝑦=1.

We first note that the second equation is not given in slope–intercept form, so we start by rearranging the equation into this form by subtracting 2π‘₯ from both sides of the equation to get 𝑦=βˆ’2π‘₯+1.

Next, we sketch both lines on the same graph. The first line has a 𝑦-intercept of 3 and a slope of 4, and the second line has a 𝑦-intercept of 1 and a slope of βˆ’2, giving us the following sketch.

We can see from the graph there is one point of intersection and so there is a unique solution to the system of equations. However, we cannot determine the exact coordinates of this point of intersection. Instead, we can note that the π‘₯-coordinate of the point of intersection lies between the lines π‘₯=βˆ’0.5 and π‘₯=0 and the 𝑦-coordinate of the point of intersection lies between the lines 𝑦=1.5 and 𝑦=2.

So, although we could not use the graphical method to determine the exact solution to this system of equations, we could determine that there was a unique solution and that the solution would satisfy the inequalities βˆ’0.5<π‘₯<0 and 1.5<𝑦<2.

This then completes the process for solving a system of linear equations graphically, and we can write the steps as follows.

How To: Solving a System of Linear Equations Graphically

  1. If we are not given the graphs of the linear equations, we should sketch the graphs of each line on the same coordinate axes. To do this, we may need to rearrange each equation into slope–intercept form.
  2. Any point where all of the lines intersect is a solution to the system of equations. In particular, if there is no such point, then there are no solutions to the system, and if the lines are coincident, then there will be an infinite number of solutions.
  3. If possible, read off the coordinates of the point of intersection to find the exact values of the solution. However, if this is not possible, use the minor grid lines to find a range in which the solutions will lie.

Let’s now see how we can apply this process to determine the solutions to a system of linear equations graphically.

Example 1: Solving a System of Two Linear Equations Using the Given Graph

Use the graph to solve the simultaneous equations 𝑦=4π‘₯βˆ’2,𝑦=βˆ’π‘₯+3.

Answer

We recall that the solution to a system of equations is given by the coordinates of the point of intersection between the graphs of all of the equations. This means the coordinates of the point of intersection between the two lines tells us the solution to the simultaneous equations.

We see that the π‘₯-coordinate of this point is 1 and the 𝑦-coordinate is 2. This tells us that π‘₯=1 and 𝑦=2 is a solution to the simultaneous equations, and we could verify this by substituting these values into the equations.

If we substitute π‘₯=1 into the first equation, we get 𝑦=4(1)βˆ’2=2, which agrees with our solution. Similarly, if we substitute π‘₯=1 into the second equation, we get 𝑦=βˆ’(1)+3=2, which also agrees with our solution. Since both equations hold true, this verifies the solution.

Since this is the only point of intersection, it is the only solution to the simultaneous equations.

Therefore, the only solution is π‘₯=1 and 𝑦=2.

In our next example, we will demonstrate how to give the solutions to a system of equations using set notation.

Example 2: Solving a System of Two Linear Equations Using Their Graph

Find the solution set of the two equations represented by 𝐿 and 𝐿.

Answer

We recall that the solution to a system of equations is given by the coordinates of the point of intersection between the graphs of all of the equations. This means the coordinates of the point of intersection between the two lines gives us the solution to this system of equations.

Since there is only one point of intersection, we can conclude there is only one solution and in turn, this means the solution set will only have one member. We can determine this unique solution by finding the coordinates of the point of intersection from the graph.

We see that the coordinates of this point are (2,3), which means that the pair of values π‘₯=2 and 𝑦=3 is the only solution to this system of equations. To write this as a solution set, we recall that we can represent solutions as an ordered pair, which is the same as its coordinates.

Therefore, the solution set of the two equations represented by 𝐿 and 𝐿 is {(2,3)}.

In our previous two examples, we were given the graphs of the linear equations. However, this will not always be the case. Let’s now see some examples where we solve a system of linear equations by first sketching the graphs of each equation.

Example 3: Solving a System of Two Linear Equations by Plotting a Graph

By plotting the graphs of 𝑦=π‘₯βˆ’1 and 𝑦=5π‘₯+7, find the point that satisfies both equations simultaneously.

Answer

We recall that if (π‘₯,𝑦) satisfies both equations simultaneously, then the point must lie on the graphs of both equations. Hence, it will be a point of intersection between both graphs. Similarly, if (π‘₯,𝑦) is a point of intersection between both graphs, then it satisfies both equations and so is a solution to the system of equations. Therefore, we can find the solutions to this system by finding the coordinates of the points of intersection. We will do this by sketching both graphs.

There are many different ways of sketching the graph of a straight line. For example, to sketch the line 𝑦=π‘šπ‘₯+𝑏, we recall that this line will have 𝑦-intercept 𝑏 and π‘₯-intercept βˆ’π‘π‘š, provided π‘šβ‰ 0 and its slope is π‘š. Connecting these points with a line allows us to sketch the line.

We can apply this to each line. First, the line 𝑦=π‘₯βˆ’1 will have a 𝑦-intercept of βˆ’1 and an π‘₯-intercept of βˆ’(βˆ’1)1=1. Second, the line 𝑦=5π‘₯+7 will have a 𝑦-intercept of 7 and an π‘₯-intercept of βˆ’75.

Since this is not an integer, we can find a different point by substitution. We substitute π‘₯=βˆ’1 into the equation to get 𝑦=5(βˆ’1)+7=2.

This gives us the following.

We can see that the lines intersect to the left of the point (βˆ’1,2). Let’s try π‘₯=βˆ’2. Substituting π‘₯=βˆ’2 into the equation of the first line gives 𝑦=βˆ’2βˆ’1=βˆ’3.

Substituting π‘₯=βˆ’2 into the equation of the second line gives 𝑦=5(βˆ’2)+7=βˆ’10+7=βˆ’3.

We note both equations give 𝑦=βˆ’3. Hence, π‘₯=βˆ’2 and 𝑦=βˆ’3 satisfies both equations and we can say that the point that satisfies both equations is (βˆ’2,βˆ’3).

Example 4: Solving a System of Two Linear Equations by Plotting a Graph

Plot the graphs of the simultaneous equations 𝑦=2π‘₯+7,𝑦=2π‘₯βˆ’4, and then solve the system.

Answer

We recall that the points of intersection of the graphs of both equations will tell us the solutions to the system of equations. This means we can solve the system by sketching both equations, since these are linear equations given in slope–intercept form, there are many ways of doing this.

For example, to sketch the line 𝑦=π‘šπ‘₯+𝑏, we recall that this line will have 𝑦-intercept 𝑏 and π‘₯-intercept βˆ’π‘π‘š, provided π‘šβ‰ 0 and its slope is π‘š. Connecting these points with a line allows us to sketch the line.

For the first line, we have that the 𝑦-intercept is at 7 and the π‘₯-intercept is βˆ’72. Since this is not an integer, we can find another point on the line. We substitute π‘₯=βˆ’1 into the equation to get 𝑦=2(βˆ’1)+7=5.

For the second line, we have that the 𝑦-intercept is at βˆ’4 and the π‘₯-intercept is at βˆ’(βˆ’4)2=2. Connecting the intercepts of each line gives us the following sketch.

We can see that these lines are parallel; this means that there are no points of intersection, so the system has no solutions.

In fact, we could have saved ourselves the effort of drawing the graphs; if we look at the equations of the two lines, we will see they are given in slope–intercept form. We know that the coefficient of π‘₯ tells us the slope of the line and we can see in both cases this is 2. Similarly, we can see that both lines have distinct 𝑦-intercepts from their equations. This tells us that these lines are parallel (they have the same slope) and they are distinct (they pass through different 𝑦-intercepts). Hence, the lines do not intersect and the system has no solutions.

Using either method, we have shown that there are no solutions as both equations represent parallel lines.

In the next example, we will demonstrate how to solve a system of linear equations when they are not given in slope–intercept form.

Example 5: Solving a System of Two Linear Equations by Plotting a Graph

By plotting the graphs of 12π‘₯βˆ’9𝑦=24 and 4π‘₯βˆ’3𝑦=8, determine the π‘₯- and 𝑦-coordinates of the point that satisfies both equations simultaneously.

Answer

We recall that the solution to a system of equations is given by the coordinates of the point of intersection between the graphs of all of the equations. This means the coordinates of the point of intersection between the two lines tell us the solution to the simultaneous equations. Therefore, we can solve this system by plotting each equation on the same graph and determining the coordinates of the points of intersection.

We will plot both graphs by finding their respective π‘₯- and 𝑦-intercepts. We substitute π‘₯=0 into the equation of the first line and solve to get 12(0)βˆ’9𝑦=24𝑦=24βˆ’9=βˆ’83.

Since this is not an integer, we will use a different point. We substitute π‘₯=βˆ’1 into the equation to get 12(βˆ’1)βˆ’9𝑦=24.

We then rearrange to get 𝑦=24+12βˆ’9=βˆ’4.

We substitute 𝑦=0 into the equation of the first line and solve to get 12π‘₯βˆ’9(0)=24π‘₯=2412=2.

We substitute π‘₯=0 into the equation of the second line and solve to get 4(0)βˆ’3𝑦=8𝑦=8βˆ’3=βˆ’83.

Since this is not an integer, we will use a different point. We substitute π‘₯=βˆ’1 into the equation to get 4(βˆ’1)βˆ’3𝑦=8.

We then rearrange to get 𝑦=8+4βˆ’3=βˆ’4.

We substitute 𝑦=0 into the equation of the second line and solve to get 4π‘₯βˆ’3(0)=8π‘₯=84=2.

So, the π‘₯- and 𝑦-intercepts of both lines are the same; hence, the lines are coincident. We can sketch the lines as follows.

Every point on the line satisfies both equations, so each point is a solution to the system of equations.

We could have also determined this from the system of equations itself. If we rearrange the equations we are given into slope–intercept form, we have 12π‘₯βˆ’9𝑦=24βˆ’9𝑦=βˆ’12π‘₯+24𝑦=βˆ’12π‘₯+24βˆ’9𝑦=43π‘₯βˆ’83 and 4π‘₯βˆ’3𝑦=8βˆ’3𝑦=βˆ’4π‘₯+8𝑦=βˆ’4π‘₯+8βˆ’3𝑦=43π‘₯βˆ’83.

We see that both equations represent the same line, so they are coincident.

Therefore, we have shown that the two lines are coincident, so there are an infinite number of solutions.

In our next example, we will use a given graph to determine the system of equations it can be used to solve.

Example 6: Identifying the Set of Simultaneous Equations That Can Be Solved Using a Given Graph

Which of the following sets of simultaneous equations could be solved using the given graph?

  1. 𝑦=2π‘₯βˆ’4, 𝑦=π‘₯+5
  2. 𝑦=βˆ’4π‘₯+2, 𝑦=5π‘₯βˆ’1
  3. 𝑦=2π‘₯βˆ’4, 𝑦=βˆ’π‘₯+5
  4. 𝑦=βˆ’4π‘₯+2, 𝑦=5π‘₯+1
  5. 𝑦=2π‘₯+4, 𝑦=βˆ’π‘₯+5

Answer

We recall that the coordinates of the point of intersection between the two lines tells us the solution to the simultaneous equations given by the equations of the lines. Therefore, a system of equations can be solved using this graph.

Hence, we need to determine the equations of both lines. We will do this by recalling that the equation of a line in slope–intercept form is 𝑦=π‘šπ‘₯+𝑏, where π‘š is the slope and 𝑏 is the 𝑦-intercept. We can determine both of these values from the diagram. First, the blue line has 𝑦-intercept 5 and the red line has 𝑦-intercept βˆ’4. Next, we determine the slope of each line by using the fact that a line passing through (π‘₯,𝑦) and (π‘₯,𝑦) has slope π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

The blue line passes through the points (0,5) and (1,4), so it has slope π‘š=4βˆ’51βˆ’0=βˆ’1.

The red line passes through the points (0,βˆ’4) and (1,βˆ’2), so it has slope π‘š=βˆ’2βˆ’(βˆ’4)1βˆ’0=2.

Hence, the blue line has π‘š=βˆ’1 and 𝑏=5 and the equation 𝑦=βˆ’π‘₯+5, and the red line has π‘š=2 and 𝑏=βˆ’4 and the equation 𝑦=2π‘₯βˆ’4. This gives us the system of equations 𝑦=βˆ’π‘₯+5,𝑦=2π‘₯βˆ’4.

The solutions to this system of linear equations, given by the coordinates of the point of intersection, are π‘₯=3 and 𝑦=2.

This is the system of linear equations in option C, 𝑦=2π‘₯βˆ’4, 𝑦=βˆ’π‘₯+5.

In our final example, we will use a given graph of a system of two linear equations to find approximate ranges in which the solution lies.

Example 7: Using a Given Graph to Find Appropriate Ranges for Solutions of Given Simultaneous Equations

Use the shown graph to find appropriate ranges for the solutions to the simultaneous equations 𝑦=3π‘₯βˆ’3,5π‘₯+7𝑦=βˆ’2.

  1. 0.5<π‘₯<0.75 and βˆ’3<𝑦<βˆ’1
  2. 0.5<π‘₯<0.75 and βˆ’1<𝑦<βˆ’0.75
  3. βˆ’1<π‘₯<βˆ’0.75 and 0.5<𝑦<0.75
  4. βˆ’0.75<π‘₯<βˆ’0.5 and 0.75<𝑦<1
  5. 0.75<π‘₯<1 and βˆ’0.75<𝑦<βˆ’0.5

Answer

We recall that the solution to a system of equations is given by the coordinates of the point of intersection between the graphs of all of the equations.

However, we cannot determine the coordinates of this point of intersection directly from the diagram, since it does not lie directly on the grid lines. We can find a range of values in which the coordinates of this point lies by using the minor grid lines.

First, we note that the grid lines split 1 into 4 equal segments, so each increment is 0.25. Hence, the point of intersection lies between the lines π‘₯=0.5 and π‘₯=0.75. We can do the same with horizontal lines.

We can see that the point of intersection lies between the lines 𝑦=βˆ’1 and 𝑦=βˆ’0.75. Hence, if (π‘₯,𝑦) is the solution to this system, then 0.5<π‘₯<0.75 and βˆ’1<𝑦<βˆ’0.75, which is option B.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • Every solution to a system of equations is a point of intersection on all of their graphs. Conversely, every point of intersection on all of the graphs of the equations will be a solution to the system of equations. Hence, we can solve systems of equations by plotting their graphs and finding the coordinates of the points of intersection.
  • If there are no points of intersection between all the graphs, then there are no solutions to the system of equations.
  • A system of linear equations can have 0, 1, or an infinite number of solutions depending on whether the lines are parallel, intersect at a unique point, or are coincident. A system with 0 solutions is called an inconsistent system, a system with 1 solution is called an independent system, and a system with an infinite number of solutions is called a dependent system.
  • We can estimate the solution to a system of linear equations graphically by finding an approximate range for the point of intersection by using minor grid lines.

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